Answer Posted / guest
#include<conio.h>
#include<stdio.h>
void main()
{
int a[10],b[10];
int c=1,n=5;
a[0]=0;a[1]=1;a[2]=0;
b[0]=0;
printf("Enter no. of rows to print ");
scanf("%d",&n);
for(int i=0;i<n;i++)
{
for(int j=n-2;j<0;j++)
printf(" ");
c++;
for(int k=0;k<c;k++)
{
b[k+1]=a[k]+a[k+1];
}
b[c]=0;
for(int k=1;k<=i+1;k++)
{
printf("%d ",a[k]);
}
printf("\n");
for(int k=0;k<i+3;k++)
a[k]=b[k];
}
getch();
}
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create the same BST or not.
For example:
Array1:10 5 20 15 30
Array2:10 20 15 30 5
Result: True
Array1:10 5 20 15 30
Array2:10 15 20 30 5
Result: False
One Approach is Pretty Clear by creating BST O(nlogn) then
checking two tree for identical O(N) overall O(nlogn) ..we
need there exist O(N) Time & O(1) Space also without extra
space .Algorithm ??
DevoCoder
guest
Posted 3 months ago #
#define true 1
#define false 0
int check(int a1[],int a2[],int n1,int n2)
{
int i;
//n1 size of array a1[] and n2 size of a2[]
if(n1!=n2) return false;
//n1 and n2 must be same
for(i=0;i
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